Question

the shortest and longest wavelength of pastchen series in hydrogen spectrum is respectively A=p/r,144/7r B= 4/r ,36/5r C=16/r,400/9r D=144/r,9/r​

Answer 1

Answer:

A: 9/R ,144/7R

Explanation:

1/lembda =R [ 1÷n'^2 - 1÷n`^2 ]

is formula for finding wavelength

so for paschan series:

n1=3 n2=4 (for longest wavelength)

1/lembda =R [ 1÷3^2 - 1÷4^2 ]

=R [ 1÷9 - 1÷16 ] taking LCM

=R [ 16-9 /144 ]

1/lembda =7R/144

so;

lembda = 144 /7R for longest wavelength

now for shortest wavelength we use n2= ∞

bcz electron coming from far will be more energetic and will emit short wavelength

so putting values n1=3 and n2= ∞

1/lembda =R [ 1÷3^2 - 1÷ ∞^2 ]

as we know that (1/ ∞ = zero)

so we get

1/lembda =R ÷9

lembda= 9/R for shortest wavelength